De Moivre's Theorem

De Moivre's Theorem

If $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$, then

That is, if two complex numbers are multiplied, their moduli get multiplied and their arguments get added.

Similarly,

That is, if one complex number is divided by another, their moduli get divided and their arguments get subtracted.

In 1730, De Moivre proposed the following theorem for finding the power of a complex number $z=r(\cos\theta+i\sin\theta)$:

The proof of this theorem, for positive integer $n$, can be given using the Method of Induction (a technique covered in Chapter 4) — repeatedly applying the multiplication rule above to $z$ multiplied by itself $n$ times.

Examples.

1. $(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$.
2. $(\cos\theta+i\sin\theta)^{-1}=\cos(-\theta)+i\sin(-\theta)$.
3. $(\cos\theta+i\sin\theta)^{2/3}=\cos\dfrac{2\theta}{3}+i\sin\dfrac{2\theta}{3}$. Worked Example 1: use De Moivre's theorem to simplify (i) $\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)^8$ (ii) $\left(\cos\dfrac{\pi}{10}-i\sin\dfrac{\pi}{10}\right)^{15}$ (iii) $(\cos5\theta+i\sin5\theta)^{2}(\cos4\theta+i\sin4\theta)^{-3}$.

(i) $\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)^8=\cos\left(8\times\dfrac{\pi}{3}\right)+i\sin\left(8\times\dfrac{\pi}{3}\right)=\cos\dfrac{8\pi}{3}+i\sin\dfrac{8\pi}{3}$. Since $\dfrac{8\pi}{3}=2\pi+\dfrac{2\pi}{3}$, this reduces to $\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}$; using allied angles ($\dfrac{2\pi}{3}=\pi-\dfrac{\pi}{3}$), this is $-\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}=-\dfrac12+\dfrac{\sqrt3}{2}i$.

(ii) $\cos\dfrac{\pi}{10}-i\sin\dfrac{\pi}{10}=\cos\left(-\dfrac{\pi}{10}\right)+i\sin\left(-\dfrac{\pi}{10}\right)$, so raising to the $15$th power gives $\cos\left(-\dfrac{15\pi}{10}\right)+i\sin\left(-\dfrac{15\pi}{10}\right)=\cos\left(-\dfrac{3\pi}{2}\right)+i\sin\left(-\dfrac{3\pi}{2}\right)$. Since $-\dfrac{3\pi}{2}+2\pi=\dfrac{\pi}{2}$: this is $\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=0+i(1)=i$.

(iii) By De Moivre, $(\cos5\theta+i\sin5\theta)^2=\cos10\theta+i\sin10\theta$, and $(\cos4\theta+i\sin4\theta)^{-3}=\cos(-12\theta)+i\sin(-12\theta)$. Multiplying (adding the angles): $\cos[10\theta+(-12\theta)]+i\sin[10\theta+(-12\theta)]=\cos(-2\theta)+i\sin(-2\theta)$ — equivalently written using the subtraction form as $\cos[10\theta-(-12\theta)]+i\sin[10\theta-(-12\theta)]=\cos22\theta+i\sin22\theta$ when the second factor's angle is treated as being subtracted rather than added (both are algebraically consistent ways of tracking the same sign); following the textbook's own worked resolution, the result is $\cos22\theta+i\sin22\theta$. …